Uniform convergence of harmonic functions to $0$ on compact subsets

Let $D \subset \mathbb{C}$ be an open, connected set and let $\{ u_n \}$ be a sequence of harmonic functions with $u_n: D \longrightarrow (0, \infty)$. Show that if $u_n(z_0) \rightarrow 0$ for some $z_0 \in D$, then $u_n \rightarrow 0$ uniformly on compact subsets of $D$.

If you could offer a hint or a helpful question to get me started I'd appreciate it.

Progress: I now see why showing it for the unit disk suffices (any compact subset in $D$ can be covered by finitely many disks each which is contained in $D$).

• Can you show this for the unit disc?
– zhw.
May 29 '15 at 21:59
• I now see why showing it for the unit disk suffices (any compact subset in $D$ can be covered by finitely many disks each which is contained in $D$). I'll work on showing it for the unit disk now. May 29 '15 at 22:10
• I meant for some. That's how the question is stated at least (it comes for a past qualifying exam). May 29 '15 at 22:21
• Harnack's inequality does it nicely for the disk.
– user147263
May 29 '15 at 23:03
• May 30 '15 at 1:38

I think you should use identity theorem for harmonic functions...

Let $D_1$ be a disc with a center at $z_0$, and $D_1\subseteq D$. For every $n$ $u_n$ has harmonic conjugate $v_n$. So we can define new sequence of analytic function $\lbrace f_n\rbrace$ on $D_1$.

$f_n = u_n + iv_n$.

Let $f = \lim_{n\to\infty}f_n$.

From Morera`s theorem $f$ is analytic on $D_1$, if, of course $D_1$ is small enough (maybe this is not so obvious).

Now define $g = \exp(f)$ and apply minimum principle to $g$. $|g(z_0)| = 1$ on $D_1$, so $f$ is constance on $D_1$, so $Ref = 0$ on $D_1$. Now you can apply identity theorem for harmonic function.

As already mentioned in the comments, one considers the case of a disk first: If $$u(z_0) \to 0$$ and $$\overline{B(z_0,r)} \subset D$$ then Harnack's inequality (see also Harnack's inequality) shows that $$u_n(z) \to 0$$ uniformly in $$\overline{B(z_0,r)}$$.

For the general case, define $$A = \{ z \in D \mid u_n(z) \to 0 \}$$ and show that both $$A$$ and $$D \setminus A$$ are open. Since $$D$$ is connected, one of the sets must be empty. $$z_0 \in A$$ implies that $$A = D$$.

So every point in $$D$$ has a neighborhood on which $$u_n(z) \to 0$$ uniformly, and that is equivalent to uniform convergence to zero on every compact subset of $$D$$.